I am stuck on this problem:
Find the equations of the tangent lines to the graph of the equation below that are parallel to the line given.
f(x) = (x+1)/(x-1)
Line : 2y+x=6
1. find the slope, $\displaystyle m$ , of the given line.
2. set $\displaystyle f'(x) = m$ and solve for values of x on the curve.
3. evaluate the corresponding y-value for each x-value found in step 2 ... then use the point-slope form of a linear equation to write the tangent line equation.
Your derivative is not correct.
$f'(x) = -\dfrac{2}{(x-1)^2}$
Hence, your numerator was incorrect. Using the correct derivative, you should be able to get the correct answers (there will be more than one).
$-\dfrac{2}{(x-1)^2} = -\dfrac{1}{2}$
$(x-1)^2 = 4$
$x-1 = \pm 2$
$x = 1 \pm 2$, so either $x = -1$ or $x = 3$.
Then, $f(-1) = 0$ and $f(3) = 2$, so both lines should have slope $-\dfrac{1}{2}$, one should pass through the point $(-1,0)$ and the other through the point $(3,2)$.